08-heaps

Top-K with a Heap

Keep only the largest k values by maintaining a small min-heap.

Algorithm

Steps

  1. Store the heap in an array.
  2. Compare parent and child indexes instead of building explicit tree nodes.
  3. Swap only when the heap order is violated.
  4. Print the deterministic final heap state for replay comparison.
bounded heap For top-k largest values, a min-heap of size k keeps the current cutoff at the root.

Complexity

  • Time: O(n log k)
  • Space: O(k)

Ruby DSA Implementation

basic.rb
Replay: real traced execution (multi-file project)
def list_string(values) = "[#{values.join(', ')}]"
def heap_insert(heap, value)
  heap << value
  child = heap.length - 1
  while child > 0
    parent = (child - 1) / 2
    break if heap[parent] <= heap[child]
    heap[parent], heap[child] = heap[child], heap[parent]
    child = parent
  end
end
def heap_pop(heap)
  smallest = heap[0]
  heap[0] = heap.pop
  parent = 0
  loop do
    left = parent * 2 + 1
    right = left + 1
    break if left >= heap.length
    child = right < heap.length && heap[right] < heap[left] ? right : left
    break if heap[parent] <= heap[child]
    heap[parent], heap[child] = heap[child], heap[parent]
    parent = child
  end
  smallest
end
heap = []
[5, 1, 9, 3, 7, 2].each do |value|
  heap_insert(heap, value)
  heap_pop(heap) if heap.length > 3
end
puts list_string(heap.sort.reverse)

  1. top-k heap ← [5]

    4child = heap.length - 15while child > 06  parent = (child - 1) / 2
    values this step[5]top-k heap5value

  2. top-k heap ← [1, 5]

    4child = heap.length - 15while child > 06  parent = (child - 1) / 2
    values this step[1, 5]top-k heap1value

  3. top-k heap ← [1, 5, 9]

    4child = heap.length - 15while child > 06  parent = (child - 1) / 2
    values this step[1, 5, 9]top-k heap9value

  4. top-k heap ← [3, 5, 9]

    4child = heap.length - 15while child > 06  parent = (child - 1) / 2
    values this step[3, 5, 9]top-k heap3value

  5. top-k heap ← [5, 7, 9]

    4child = heap.length - 15while child > 06  parent = (child - 1) / 2
    values this step[5, 7, 9]top-k heap7value

  6. top-k heap ← [5, 7, 9]

    4child = heap.length - 15while child > 06  parent = (child - 1) / 2
    values this step[5, 7, 9]top-k heap2value

  7. stdout ← [9, 7, 5]

    1def list_string(values) = "[#{values.join(', ')}]"2def heap_insert(heap, value)
    values this step[9, 7, 5]stdout[5, 7, 9]top-k heap

Implementation notes

  • The working heap is a Ruby Array used as a min-heap, starting from heap = [].
  • This page reuses heap_insert and heap_pop; both mutate the same heap array in place.
  • The input values are fixed as [5, 1, 9, 3, 7, 2], and the size bound is the literal 3.
  • Each value is inserted first with heap_insert(heap, value).
  • Immediately after insertion, heap_pop(heap) if heap.length > 3 removes the current minimum whenever the heap grows past the top-k size.
  • That means small values can enter briefly; the trace for value 2 ends back at [5, 7, 9] after trimming.
  • The replayed heap states are [5], [1, 5], [1, 5, 9], [3, 5, 9], [5, 7, 9], then [5, 7, 9].
  • The heap array itself is not descending sorted, so output uses heap.sort.reverse only for display.
  • list_string(heap.sort.reverse) prints the final top values as [9, 7, 5].

Output

[9, 7, 5]